# limit of $\lim_{x \to 2} (x-4)$

I have just begun to learn limits and I'm having some trouble understanding the subtraction law. I understand how to find the limit if it is something like $\lim_{x \to 2} (x+4)-(x+5)$ we just take the limits of both of the functions and then subtract them but when I do it for something like $\lim_{x \to 2} (x-4)$ I'm getting the wrong answer. What I am doing is, I am taking the limit of $x$ and then taking the limit of $-4$ and then subtracting it. The limit of $x$ is 2 and the limit of $-4$ is $-4$ but if I subtract them then I get $2-(-4) = 6$ which is incorrect. Should I be doing $2 - 4$ and negating the negative sign of $-4$?

Thanks

-
When $x$ is ridiculously close to $2$ (but not equal to $2$), what is $x-4$ ridiculously close to? – André Nicolas Feb 1 '13 at 18:21
Your confusion does not come from limits, but from algebra; $x-4=x+(-4)$, not $x-(-4)$. – 1015 Feb 1 '13 at 18:38

$\lim (x-4)=\lim x-\lim 4=2-4=-2$. In words, the limit of the difference is the difference of the limits if the latter limits exist.

Alternatively, $\lim (x-4)=\lim (x+(-4))=\lim x+\lim (-4)=2+(-4)=-2$. In words, the limit of the sum is the sum of the limits if the latter limits exist.

Your confusion arises because the negative sign has been represented twice.

-
OR $\lim(x-4) = \lim x + \lim(-4) = 2+(-4) = -2$. – GEdgar Feb 1 '13 at 18:24
So basically, the subtraction sign between the two remains the the subtraction sign and does not effect either x or 4 but suppose it was $(x-(-4))$ then I would correct in saying that $\lim x - \lim -4$? – Jeel Shah Feb 1 '13 at 18:26

It is quite simple in this case actually. When a function $f(x)$ is continuous (like any polynomial such as $f(x)=x-4$) then by definition this means that $\lim\limits_{x \to a}f(x)=f(a)$ for every possible $a$ value. In this case that would mean $\lim\limits_{x\to 2}(x-4)=(2-4)=-2$.

-